The function $f(x)=\log _2(-x-2)-1$ is graphed below. Plot all lattice points of the inverse. Use the labeled points as your guide.

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05-11-2026
Mathematics

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The function $f(x)=\log _2(-x-2)-1$ is graphed below. Plot all lattice points of the inverse. Use the labeled points as your guide.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-05T22:53:50-05:00

Determine the inverse function: $f^{-1}(x) = -2^{x+1} - 2$.
Recognize that $y$ is an integer only when $x+1 \geq 0$, meaning $x \geq -1$.
Substitute integer values of $x \geq -1$ into the inverse function to generate lattice points $(x, y)$.
The lattice points on the inverse function are $\boxed{(-1, -3), (0, -4), (1, -6), (2, -10), (3, -18), \dots}$.

Explanation

Analysis of the Problem

The goal is to find all lattice points (pairs of integers $(x, y)$) that lie on the graph of the inverse function $f^{-1}(x)$, where the original function is $f(x) = \log_2(-x - 2) - 1$. A lattice point $(x, y)$ is on the graph of $f^{-1}(x)$ if and only if the point $(y, x)$ is on the graph of the original function $f(x)$. This means we need to identify integer pairs $(x, y)$ such that $f(y) = x$.

Finding the Inverse Function

To find the inverse function $f^{-1}(x)$, we set $y = f(x)$ and solve for $x$ in terms of $y$:
\[y = \log_2(-x - 2) - 1\]
Adding $1$ to both sides:
\[y + 1 = \log_2(-x - 2)\]
Exponentiating both sides with base $2$:
\[2^{y+1} = -x - 2\]
Solving for $x$:
\[x = -2^{y+1} - 2\]
Thus, the inverse function is $f^{-1}(y) = -2^{y+1} - 2$. Since we want to express this in the standard notation $f^{-1}(x)$, we replace $y$ with $x$:
\[f^{-1}(x) = -2^{x+1} - 2\]
A lattice point $(x, y)$ must satisfy $y = -2^{x+1} - 2$, where both $x$ and $y$ are integers.

Identifying Lattice Points

We need $x$ and $y = -2^{x+1} - 2$ to be integers.
For $y$ to be an integer, $2^{x+1}$ must be an integer. This occurs if and only if $x + 1 \geq 0$, or $x \geq -1$. Since $x$ is an integer, the possible values for $x$ are $-1, 0, 1, 2, 3, \dots$.
Let's calculate the corresponding $y$ values for these $x$ values:

For $x = -1$: $y = -2^{-1+1} - 2 = -2^0 - 2 = -1 - 2 = -3$. Point: $(-1, -3)$.
For $x = 0$: $y = -2^{0+1} - 2 = -2^1 - 2 = -2 - 2 = -4$. Point: $(0, -4)$.
For $x = 1$: $y = -2^{1+1} - 2 = -2^2 - 2 = -4 - 2 = -6$. Point: $(1, -6)$.
For $x = 2$: $y = -2^{2+1} - 2 = -2^3 - 2 = -8 - 2 = -10$. Point: $(2, -10)$.
For $x = 3$: $y = -2^{3+1} - 2 = -2^4 - 2 = -16 - 2 = -18$. Point: $(3, -18)$.

The general form of these points is $(x, -2^{x+1} - 2)$ for all $x \in \{-1, 0, 1, 2, \dots\}$.

Conclusion

The lattice points on the graph of $f^{-1}(x) = -2^{x+1} - 2$ are given by the set of ordered pairs:
\[\{(-1, -3), (0, -4), (1, -6), (2, -10), (3, -18), \dots\}\]
These can be expressed as $\boxed{(-1, -3), (0, -4), (1, -6), (2, -10), (3, -18), \dots}$.

Examples

Lattice points are essential in digital geometry, where they represent pixelated versions of curves on a screen. If you have a logarithmic growth curve, identifying the lattice points helps computer graphic designers plot the curve accurately on a grid of pixels. For example, in a game, if a path follows a specific logarithmic function, finding the closest integer coordinates ensures the character moves along the correct pixel-based trajectory.